4
$\begingroup$

Here is some background on my question. The Weierstraß function (as well as its translates) $\wp(x)$ solves the implicit first-order ODE $$(y')^2=4y^3-g_2y-g_3.$$ Differentiating gives the second-order explicit ODE $$y''=12y^2-g_2, $$ in which $g_3$ makes no appearance.

$\wp$ can be used to express the solution of any first-order equation of the form $$(y')^2=a_3 y^3+a_2 y^2+a_1 y+a_0, $$ or equivalently, any second-order equation of the form $$y''=b_2 y^2+b_1 y+b_0. $$ In both cases, all it takes is a simple change of variables to remove one of the powers, and get the leading coefficient right.

My question is about differential equations whose RHS is one degree larger, that is $$(y')^2=a_4 y^4+a_3y^3+a_2y^2+a_1 y+a_0 , \tag{1}$$ or $$y''=b_3 y^3+b_2y^2+b_1 y+b_0. \tag{2} $$

In some special cases, such as $$y''=2k^2 y^3-(1+k^2) y, $$ and $$y''=-2y^3+(2-k^2)y,$$ the Jacobi elliptic functions $\operatorname{sn},\operatorname{dn}$ play a role in the solution.

Here is my question: For which values of the coefficients in $(1)$ and $(2)$ can the ODEs be solved explicitly in terms of special functions? What are the solutions in those cases?

Thank you!

$\endgroup$
2
$\begingroup$

The change of the function $y= \frac{\alpha z+\beta}{\gamma z+\delta}$ with $\alpha\delta-\beta\gamma=1$ transforms (1) into $$\left(z'\right)^2=a_4\left(\alpha z+\beta\right)^4+ a_3\left(\alpha z+\beta\right)^3\left(\gamma z+\delta\right) +a_2\left(\alpha z+\beta\right)^2\left(\gamma z+\delta\right)^2+ +a_1\left(\alpha z+\beta\right)\left(\gamma z+\delta\right)^3+ a_0\left(\gamma z+\delta\right)^4=\operatorname{poly}_4\left( z\right).$$ We can make the coefficient of $z^4$ disappear by the appropriate choice of $\alpha/\gamma$. The problem is thus completely equivalent to the previous differential equation with cubic right hand side and can be solved in terms of Weierstrass $\wp$-function.

The change of variable above is exactly the one which transforms the integral $\int\frac{dy}{\sqrt{\operatorname{poly}_4\left( y\right)}}$ into $\int\frac{dz}{\sqrt{\operatorname{poly}_3\left( z\right)}}$. This has the following geometric interpretation: $w^2=\operatorname{poly}_4\left( y\right)$ is an elliptic curve realized as a $2$-sheeted covering of the Riemann sphere having $4$ branch points. Cubic Weierstrass form of the curve corresponds to moving one of the branch points to $\infty$ by a homography.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.